Symmetric Difference Events Are Measurable
Suppose that a probability space \(\left( \Omega,\mathcal{F},P\right) \)is given. That is, the family of events \(\mathcal{F}\subset2^{\Omega}\)satisfies the axioms of a measurable space \begin{align*}
& \left( M_{1}\right) \ \mathcal{F}\not =\phi\\
& \left( M_{2}\right) \ \forall A\in\mathcal{F}:A^{c}\in\mathcal{F}\\
& \left( M_{3}\right) \ \forall\left\{ A_{n}\right\} _{n\in\mathbb{N} }\subset\mathcal{F}:\bigcup_{n\in\mathbb{N} }A_{n}\in\mathcal{F},
\end{align*}and the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) satisfies the axioms of probability \begin{align*}
& \left( P_{1}\right) \ \forall A\in\mathcal{F}:P\left( A\right) \geq0\\
& \left( P_{2}\right) \ P\left( \Omega\right) =1\\
& \left( P_{3}\right) \ \forall\text{mutually disjoint }\left\{
A_{n}\right\} _{n\in\mathbb{N} }\subset\mathcal{F}:P\left( \bigcup_{n\in\mathbb{N} }A_{n}\right) =\sum_{n\in\mathbb{N} }P\left( A_{n}\right) .
\end{align*}
Given a probability space \(\left( \Omega,\mathcal{F},P\right) \), what can be said about the probability of a symmetric difference event? Before answering this question, we first verify that symmetric difference events are measurable.
Proposition (Symmetric Difference Events Are Measurable)
In a probability space \(\left( \Omega,\mathcal{F},P\right) \), for arbitrary events \(A,B\in\mathcal{F}\), \[
A\Delta B\in\mathcal{F}.
\]Thus, \(\mathcal{F}\) is closed under symmetric difference.
A\Delta B\in\mathcal{F}.
\]Thus, \(\mathcal{F}\) is closed under symmetric difference.
Probability of a Symmetric Difference Event
Since the event space \(\mathcal{F}\) is closed under symmetric difference, for arbitrary events \(A,B\in\mathcal{F}\), \[
A\Delta B\in\mathcal{F}.
\]Therefore, the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) assigns the probability \(P\left( A\Delta B\right) \) to this event, and \[
P\left( A\Delta B\right) =P\left( A\right) +P\left( B\right) -2P\left(
A\cap B\right) .
\]
Proposition (Probability of a Symmetric Difference Event)
In a probability space \(\left( \Omega,\mathcal{F},P\right) \), for arbitrary events \(A,B\in\mathcal{F}\), \[
P\left( A\Delta B\right) =P\left( A\right) +P\left( B\right) -2P\left(
A\cap B\right) .
\]
P\left( A\Delta B\right) =P\left( A\right) +P\left( B\right) -2P\left(
A\cap B\right) .
\]
Example (Symmetric Difference with the Empty Event)
For an arbitrary event \(A\in\mathcal{F}\), its symmetric difference with the empty event \(\phi \in\mathcal{F}\) satisfies \begin{align*}
A\Delta\phi & =\left( A\backslash\phi\right) \cup\left( \phi\backslash A\right) \\
& =A\cup\phi\\
& =A.
\end{align*}Therefore,
\[
P\left( A\Delta\phi\right) =P\left( A\right) .
\]The same result can be derived from the preceding proposition:
\begin{align*}
P\left( A\Delta\phi\right) & =P\left( A\right) +P\left( \phi\right)
-2P\left( A\cap\phi\right) \\
& =P\left( A\right) +P\left( \phi\right) -2P\left( \phi\right)
\quad\because A\cap\phi=\phi\\
& =P\left( A\right) +0-0\quad\because P\left( \phi\right) =0\\
& =P\left( A\right) .
\end{align*}Thus, the same result is obtained.
A\Delta\phi & =\left( A\backslash\phi\right) \cup\left( \phi\backslash A\right) \\
& =A\cup\phi\\
& =A.
\end{align*}Therefore,
\[
P\left( A\Delta\phi\right) =P\left( A\right) .
\]The same result can be derived from the preceding proposition:
\begin{align*}
P\left( A\Delta\phi\right) & =P\left( A\right) +P\left( \phi\right)
-2P\left( A\cap\phi\right) \\
& =P\left( A\right) +P\left( \phi\right) -2P\left( \phi\right)
\quad\because A\cap\phi=\phi\\
& =P\left( A\right) +0-0\quad\because P\left( \phi\right) =0\\
& =P\left( A\right) .
\end{align*}Thus, the same result is obtained.
Example (When One Event Is Contained in the Other)
If events \(A,B\in\mathcal{F}\) satisfy \(B\subset A\), then \begin{align*}
A\Delta B & =\left( A\backslash B\right) \cup\left( B\backslash A\right)
\\
& =\left( A\backslash B\right) \cup\phi\\
& =A\backslash B.
\end{align*}Therefore,
\begin{align*}
P\left( A\Delta B\right) & =P\left( A\backslash B\right) \\
& =P\left( A\right) -P\left( B\right) \quad\because B\subset A.
\end{align*}The same result follows from the preceding proposition:
\begin{align*}
P\left( A\Delta B\right) & =P\left( A\right) +P\left( B\right)
-2P\left( A\cap B\right) \\
& =P\left( A\right) +P\left( B\right) -2P\left( B\right) \quad\because B\subset A\\
& =P\left( A\right) -P\left( B\right) .
\end{align*}Thus, the same result is obtained.
A\Delta B & =\left( A\backslash B\right) \cup\left( B\backslash A\right)
\\
& =\left( A\backslash B\right) \cup\phi\\
& =A\backslash B.
\end{align*}Therefore,
\begin{align*}
P\left( A\Delta B\right) & =P\left( A\backslash B\right) \\
& =P\left( A\right) -P\left( B\right) \quad\because B\subset A.
\end{align*}The same result follows from the preceding proposition:
\begin{align*}
P\left( A\Delta B\right) & =P\left( A\right) +P\left( B\right)
-2P\left( A\cap B\right) \\
& =P\left( A\right) +P\left( B\right) -2P\left( B\right) \quad\because B\subset A\\
& =P\left( A\right) -P\left( B\right) .
\end{align*}Thus, the same result is obtained.
Example (When the Events Are Mutually Disjoint)
If events \(A,B\in\mathcal{F}\) are mutually disjoint, then \begin{align*}
A\Delta B & =\left( A\backslash B\right) \cup\left( B\backslash A\right)
\\
& =A\cup B\quad\because A\cap B=\phi.
\end{align*}Therefore,
\begin{align*}
P\left( A\Delta B\right) & =P\left( A\cup B\right) \\
& =P\left( A\right) +P\left( B\right) -P\left( A\cap B\right) \\
& =P\left( A\right) +P\left( B\right) -0\quad\because A\cap B=\phi\\
& =P\left( A\right) +P\left( B\right) .
\end{align*}The same result follows from the preceding proposition:
\begin{align*}
P\left( A\Delta B\right) & =P\left( A\right) +P\left( B\right)
-2P\left( A\cap B\right) \\
& =P\left( A\right) +P\left( B\right) -2\cdot0\quad\because A\cap B=\phi\\
& =P\left( A\right) +P\left( B\right) .
\end{align*}Thus, the same result is obtained.
A\Delta B & =\left( A\backslash B\right) \cup\left( B\backslash A\right)
\\
& =A\cup B\quad\because A\cap B=\phi.
\end{align*}Therefore,
\begin{align*}
P\left( A\Delta B\right) & =P\left( A\cup B\right) \\
& =P\left( A\right) +P\left( B\right) -P\left( A\cap B\right) \\
& =P\left( A\right) +P\left( B\right) -0\quad\because A\cap B=\phi\\
& =P\left( A\right) +P\left( B\right) .
\end{align*}The same result follows from the preceding proposition:
\begin{align*}
P\left( A\Delta B\right) & =P\left( A\right) +P\left( B\right)
-2P\left( A\cap B\right) \\
& =P\left( A\right) +P\left( B\right) -2\cdot0\quad\because A\cap B=\phi\\
& =P\left( A\right) +P\left( B\right) .
\end{align*}Thus, the same result is obtained.
Exercises
Problem (Probability of a Symmetric Difference Event)
In a probability space \(\left( \Omega,\mathcal{F},P\right) \), suppose that events \(A,B\in\mathcal{F}\) satisfy
\begin{align*}
P\left( A\right) & =\frac{2}{3}\\
P\left( B\right) & =\frac{1}{2}\\
P\left( A\cap B\right) & =\frac{1}{4}.
\end{align*}Find
\[
P\left( A\Delta B\right) .
\]
\begin{align*}
P\left( A\right) & =\frac{2}{3}\\
P\left( B\right) & =\frac{1}{2}\\
P\left( A\cap B\right) & =\frac{1}{4}.
\end{align*}Find
\[
P\left( A\Delta B\right) .
\]
Problem (Symmetric Difference of Mutually Disjoint Events)
In a probability space \(\left( \Omega,\mathcal{F},P\right) \), suppose that events \(A,B\in \mathcal{F}\) are mutually disjoint and satisfy \begin{align*}
P\left( A\right) & =\frac{3}{10}\\
P\left( B\right) & =\frac{2}{5}.
\end{align*}Find
\[
P\left( A\Delta B\right) .
\]
P\left( A\right) & =\frac{3}{10}\\
P\left( B\right) & =\frac{2}{5}.
\end{align*}Find
\[
P\left( A\Delta B\right) .
\]
Problem (Symmetric Difference under Set Inclusion)
In a probability space \(\left( \Omega,\mathcal{F},P\right) \), suppose that events \(A,B\in\mathcal{F}\) satisfy
\[
B\subset A
\]and
\begin{align*}
P\left( A\right) & =\frac{4}{5}\\
P\left( B\right) & =\frac{1}{3}.
\end{align*}Find
\[
P\left( A\Delta B\right) .
\]
\[
B\subset A
\]and
\begin{align*}
P\left( A\right) & =\frac{4}{5}\\
P\left( B\right) & =\frac{1}{3}.
\end{align*}Find
\[
P\left( A\Delta B\right) .
\]
Problem (Finding an Intersection Probability from a Symmetric Difference Probability)
In a probability space \(\left( \Omega,\mathcal{F},P\right) \), suppose that events \(A,B\in\mathcal{F}\) satisfy \begin{align*}
P\left( A\right) & =\frac{3}{4}\\
P\left( B\right) & =\frac{2}{3}\\
P\left( A\Delta B\right) & =\frac{5}{12}.
\end{align*}Find
\[
P\left( A\cap B\right) .
\]
P\left( A\right) & =\frac{3}{4}\\
P\left( B\right) & =\frac{2}{3}\\
P\left( A\Delta B\right) & =\frac{5}{12}.
\end{align*}Find
\[
P\left( A\cap B\right) .
\]
Problem (Symmetric Difference and Union Event)
In a probability space \(\left( \Omega,\mathcal{F},P\right) \), suppose that events \(A,B\in\mathcal{F}\) satisfy
\begin{align*}
P\left( A\cup B\right) & =\frac{5}{6}\\
P\left( A\cap B\right) & =\frac{1}{3}.
\end{align*}Find
\[
P\left( A\Delta B\right) .
\]
\begin{align*}
P\left( A\cup B\right) & =\frac{5}{6}\\
P\left( A\cap B\right) & =\frac{1}{3}.
\end{align*}Find
\[
P\left( A\Delta B\right) .
\]
Problem (Probability of a Symmetric Difference Event)
Consider the experiment of rolling one die and observing the outcome. Its sample space is
\[
\Omega=\left\{ 1,2,3,4,5,6\right\} .
\]Let the event space be
\[
\mathcal{F}=2^{\Omega},
\]and suppose that the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) assigns to each event \(A\in\mathcal{F}\) the probability \[
P\left( A\right) =\frac{\left\vert A\right\vert }{\left\vert \Omega
\right\vert }=\frac{\left\vert A\right\vert }{6}.
\]Then \(\left( \Omega,\mathcal{F},P\right) \) is a probability space. Find the probability of the symmetric difference between the event that an odd number is rolled and the event that a number greater than or equal to \(4\) is rolled.
\[
\Omega=\left\{ 1,2,3,4,5,6\right\} .
\]Let the event space be
\[
\mathcal{F}=2^{\Omega},
\]and suppose that the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) assigns to each event \(A\in\mathcal{F}\) the probability \[
P\left( A\right) =\frac{\left\vert A\right\vert }{\left\vert \Omega
\right\vert }=\frac{\left\vert A\right\vert }{6}.
\]Then \(\left( \Omega,\mathcal{F},P\right) \) is a probability space. Find the probability of the symmetric difference between the event that an odd number is rolled and the event that a number greater than or equal to \(4\) is rolled.
Problem (Probability of a Symmetric Difference Event)
Consider the experiment of rolling one die twice and observing the two outcomes. Its sample space is
\[
\Omega=\left\{ \left( i,j\right) \ |\ i,j\in\left\{ 1,2,3,4,5,6\right\}
\right\} .
\]Let the event space be
\[
\mathcal{F}=2^{\Omega},
\]and suppose that the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) assigns to each event \(A\in\mathcal{F}\) the probability \[
P\left( A\right) =\frac{\left\vert A\right\vert }{\left\vert \Omega
\right\vert }=\frac{\left\vert A\right\vert }{36}.
\]Then \(\left( \Omega,\mathcal{F},P\right) \) is a probability space. Find the probability of the symmetric difference between the event that the first roll is even and the event that the second roll is even.
\[
\Omega=\left\{ \left( i,j\right) \ |\ i,j\in\left\{ 1,2,3,4,5,6\right\}
\right\} .
\]Let the event space be
\[
\mathcal{F}=2^{\Omega},
\]and suppose that the probability measure \(P:\mathcal{F}\rightarrow\mathbb{R} \) assigns to each event \(A\in\mathcal{F}\) the probability \[
P\left( A\right) =\frac{\left\vert A\right\vert }{\left\vert \Omega
\right\vert }=\frac{\left\vert A\right\vert }{36}.
\]Then \(\left( \Omega,\mathcal{F},P\right) \) is a probability space. Find the probability of the symmetric difference between the event that the first roll is even and the event that the second roll is even.
Problem (Associativity of Symmetric Difference)
In a probability space \(\left( \Omega,\mathcal{F},P\right) \), suppose that events \(A,B,C\in\mathcal{F} \) satisfy
\begin{align*}
P\left( A\right) & =\frac{1}{2}\\
P\left( B\right) & =\frac{1}{2}\\
P\left( C\right) & =\frac{1}{2}\\
P\left( A\cap B\right) & =\frac{1}{4}\\
P\left( A\cap C\right) & =\frac{1}{4}\\
P\left( B\cap C\right) & =\frac{1}{4}\\
P\left( A\cap B\cap C\right) & =\frac{1}{8}.
\end{align*}Find
\[
P\left( \left( A\Delta B\right) \Delta C\right) .
\]
\begin{align*}
P\left( A\right) & =\frac{1}{2}\\
P\left( B\right) & =\frac{1}{2}\\
P\left( C\right) & =\frac{1}{2}\\
P\left( A\cap B\right) & =\frac{1}{4}\\
P\left( A\cap C\right) & =\frac{1}{4}\\
P\left( B\cap C\right) & =\frac{1}{4}\\
P\left( A\cap B\cap C\right) & =\frac{1}{8}.
\end{align*}Find
\[
P\left( \left( A\Delta B\right) \Delta C\right) .
\]
Problem (When the Probability of a Symmetric Difference Is \(0\))
In a probability space \(\left( \Omega,\mathcal{F},P\right) \), suppose that events \(A,B\in \mathcal{F}\) satisfy \[
P\left( A\Delta B\right) =0.
\]Show that
\[
P\left( A\right) =P\left( B\right) .
\]
P\left( A\Delta B\right) =0.
\]Show that
\[
P\left( A\right) =P\left( B\right) .
\]
